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Prove the sequence $\{a_i\}$ defined by $a_1=1 \text{ and } a_{i+1} = 1 + \frac{1}{a_i}$ is Cauchy. And prove it converges to $\sqrt{2}$. I want to show $\lim\limits_{n\to\infty}(a_{i+1}-a_i)=0$ for the first part, but I don't know how to approach it.

Martin Sleziak
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PPDS
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  • Probably you need to study better the definition of a Cauchy sequence. Showing the limit you wrote is a necessary, but not sufficient, condition for the sequence to be Cauchy. – Giovanni Mascellani Oct 16 '14 at 20:17
  • Does it really converge to the square root of 2? Plugging in couple of values, starting at 1, gives me 1.6 , 1.625 , 1.61538 etc which actually seems to converge to another well known constant – imranfat Oct 16 '14 at 20:17
  • It's easy to show that it goes to the golden ratio assuming that $\displaystyle \exists L\in\mathbb{R} \ s.t \ \lim_{n\to\infty}a_n=L$ – Galc127 Oct 16 '14 at 20:18
  • @Galc127 That's what I though too – imranfat Oct 16 '14 at 20:20
  • Related: http://math.stackexchange.com/questions/315366/continued-fraction-1-1-1 – Martin Sleziak Oct 17 '14 at 13:07

4 Answers4

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$\lim_{n \to \infty} (a_{i+1} - a_i) = 0$ isn't a sufficient condition for $\{a_i\}$ to be Cauchy. What is a sufficient condition is for there to exist constants $C > 0$ and $0 < r < 1$ such that $|a_{i+1} - a_i| \le C r^i$.

A quick induction shows that $\dfrac 32 \le a_i \le 2$ for all $i \ge 2$. This is because $a_2 =2$ and $$ \frac 32 \le a_i \le 2 \implies \frac 12 \le \frac 1{a_i} \le \frac 23 \implies \frac 32 \le 1 + \frac{1}{a_i} \le \frac 53 \implies \frac 32 \le a_{i+1} \le 2.$$ This implies, if $i \ge 2$, that $$ |a_{i+2} - a_{i+1}| = \left| \frac{1}{a_{i+1}} - \frac{1}{a_i} \right| = \frac{1}{|a_{i+1}||a_i|} |a_{i+1} - a_i| \le \frac 49 |a_{i+1} - a_i|.$$ In particular, if $i \ge 2$, another induction argument easily shows $$|a_{i+1} - a_i| \le \left( \frac 49 \right)^{i-2} |a_3 - a_2| = \frac{81}{32} \left( \frac 49 \right)^i$$ because $|a_3 - a_2| = \dfrac 12$.

Umberto P.
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It actually converges to $\phi = \frac{\sqrt5 +1}2$. See that said number is a fixed point of the iteration ($\phi = 1 + \frac1{\phi}$) To establish convergence, you could use the continued fraction expansion of $\phi$ to see that $a_i$ is the $i$-th convergent of $\phi$ since $$\phi = [1; \bar1] = [1; 1\ 1\ 1 \ldots ] = 1+\frac1{1+\frac1{1+\frac1{1+\ldots}}}$$

AlexR
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Prove by induction that for any $i\ge0$ $$ a_{2i+1} \le a_{2i+3}\le a_{2i+4 }\le a_{2i+2 } .$$ That's easy, because $a_{k+2}=f(a_k)$ for some increasing function $f$, which is $$f(x)=1+\frac{1}{1+\frac{1}{x}}=2-\frac{ 1}{x+1}.$$

So $a_{2i+1}$ is increasing, $a_{2i }$ is decreasing and both converge in $[1,2]$; the (a priori different) limits solve $1\le x=f(x)\le 2$, whence they coincide and are the golden number.

Pietro Majer
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I would use this:

prove by induction that 1 $\leq a_n \leq 2$ (easy)

Then Prove that the sequence has only one limit point (two will come out of the equation : $x^2 -x -1 =0$ but your sequence is always $\geq1$ so that only makes one)

Then there is a very interesting property regarding real number sequences (could be generalized..) : If ($a_n$) is bounded and has only one limit point, then it converges to its only limit point. To prove this use bolzano-weirstrauss theorem, and suppose there is an infinity of term that are distant of the limit point, from which you will be able to extract (that's the theorem above) a converging sequence and thus a second and distinct limit point.

Here your sequence verifies the two hypothesis, so it converges -> ϕ

mvggz
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